1.4 Hutchings' theory of integration

We have so far found that if V is a type-n invariant, then $W=\partial^nV$ is a linear functional on ${\mathcal A}_n$. A question arises whether every linear functional on ${\mathcal A}_n$ arises in this way. At least if the ground ring is extended to ${\mathbb Q}$, the answer is positive:

Theorem 2 (The Fundamental Theorem of Finite Type Invariants, Kontsevich [Ko1])   Over ${\mathbb Q}$, for every $W\in{\mathcal A}^\star_n$ there exists a type ninvariant V with $W=\partial^nV$. In other words, every once-integrable weight system is fully integrable.

The problem with the Fundamental Theorem is that all the proofs we have for it are somehow ``transcendental'', using notions from realms outside the present one, and none of the known proofs settles the question over the integers (see [BS]). In this section we describe what appears to be the most natural and oldest approach to the proof, having been mentioned already in [Va1,BL]. Presently, we are stuck and the so-called ``topological'' approach does not lead to a proof. But it seems to me that it's worth studying further; when something natural fails, there ought to be a natural reason for that, and it would be nice to know what it is.

The idea of the topological approach is simple: To get from W to V, we need to ``integrate'' n times. Let's do this one integral at a time. By the definition of ${\mathcal A}_n$, we know that we can integrate once and find $W^1\in{\mathcal K}_{n-1}^\star$ so that $\partial W^1=W$. Can we work a bit harder, and find a ``good'' W¹, so that there would be a $W^2\in{\mathcal K}_{n-2}^\star$ with $\partial W^2=W^1$? Proceeding like that and assuming that all goes well along the way, we would end with a $V=W^n\in{\mathcal K}_0^\star$ with $\partial^nV=W$, as required. Thus we are naturally lead to the following conjecture, which implies the Fundamental Theorem by the backward-inductive argument just sketched:

Conjecture 1   Every once-integrable invariant of n-singular knots also twice integrable. Glancing at (2), we see that this is the same as saying that $(\ker\delta^2)/(\ker\delta)=0$.

This conjecture is somewhat stronger than Theorem 2. Indeed, Theorem 2 is equivalent to Conjecture 1 restricted to the case when the given invariant has some (possibly high) derivative identically equal to 0(exercise!). But it is hard to imagine a topological proof of the restricted form of Conjecture 1 that would not prove it in full.

The difficulty in Conjecture 1 is that it's hard to say much about $\ker\delta^2$. In [Hu], Michael Hutchings was able to translate the statement $(\ker\delta^2)/(\ker\delta)=0$ to an easier-looking combinatorial-topological statement, which is implied by and perhaps equivalent to an even simpler fully combinatorial statement. Furthermore, Hutchings proved the fully combinatorial statement in the analogous case of finite type braid invariants, thus proving Conjecture 1 and Theorem 2 (over ${\mathbb Z}$) in that case, and thus proving the viability of his technique.

Hutchings' first step was to write a chain of isomorphisms reducing $(\ker\delta^2)/(\ker\delta)$ to something more manageable. Our next step will be to introduce all the spaces participating in Hutchings' chain. First, let us consider the space of all T4T relations:

Definition 1.8   Let ${\mathcal K}^1_n$ be the ${\mathbb Z}$-module generated by all (framed) knots having n-2 singularities as in Definition 1.1, and plus one additional ``Topological Relator'' singularity that locally looks like the image in Figure 6, modulo the same co-differentiability relations as in Definition 1.2. Define $\delta:{\mathcal K}^1_{n+1}\to{\mathcal K}^1_n$in the same way as for knots, using equation (1). Finally, define $b:{\mathcal K}^1_n\to{\mathcal K}_n$ by mapping the topological relator to the topological 4-term relation, the 4-term alternating sum inside the paranthesis in Figure 4.

  

Figure 6: The ``Topological Relator'' singularity.

The spaces ${\mathcal K}^1_n$ form a ladder similar to the one in (2), and, in fact, they combine with the ladder in (2) to a single commutative diagram:

$${ \begin{array}{ccccccc} \ldots\stackrel{\delta}{\longrig... ...-1} & \stackrel{\delta}{\longrightarrow}\ldots, \end{array}}$$ (3)

In this language, Stanford's theorem (Theorem 1) says that all L shapes in the above diagram (compositions $\delta\circ b$ of ``down'' followed by ``right'') are exact.

Just like singular knots had symbols which were simplar combinatorial objects (chord diagrams), so do toplogical relators have combinatorial symbols:

Definition 1.9   Let $\bar{\mathcal K}^1_n:={\mathcal K}^1_n/\delta{\mathcal K}^1_{n+1}$, and let $\pi:{\mathcal K}^1_n\to\bar{\mathcal K}^1_n$ be the projection map.

The following proposition is proved along the same lines as the standard proof of Proposition 1.6.

Proposition 1.10   $\bar{\mathcal K}^1_n$ is canonically isomorphic to the space spanned by all ``relator symbols'', chord diagrams with n-2 chords and one $\setlength{\unitlength}{0.3\standardunitlength} \begin{array}{c} \hspace{-1... ... \raisebox{-2pt}{ \input draws/BareRelator.tex } \hspace{-1.9mm} \end{array}$ piece corresponding to the special singularity of Definition 1.8. An example appears in Figure 7.

  

Figure 7: A ``relator symbol''.

We need to display one additional commutative diagram before we can come to Hutchings' chain of isomorphisms:

$$\begin{array}{ccccccc} {\mathcal K}^1_n & \stackrel{\delt... ... \longrightarrow & 0 \end{array} \qquad\text{(exact rows)}.$$ (4)

In this diagram, $\bar{b}$ is the ``symbol level'' version of b, and is induced by $b:{\mathcal K}^1_{n-1}\to{\mathcal K}_{n-1}$ in the usual manner. It can be described combinatorially by

$$\bar{b}: \if ny \smash{\makebox[0pt]{\hspace{-0.5in} \rais... ...{-8pt}{ \input draws/barb.tex } \hspace{-1.9mm} \end{array}.$$

Hutchings' chain of isomorphisms is the following chain of equalities and maps: (here the symbol $\cup$ means that the space below is a subspace of the space above, and the symbol ${ \setlength{\unitlength}{0.0004in} \begin{picture} (474,336)(0,-10) \put(2... ...87,159)(87,309) \path(387,309)(387,159) \path(12,159)(462,159) \end{picture}}$ means that the space below is a sub-quotient of the space above)

$$\begin{array}{ccccccccccccc} {\mathcal K}_n & & {\mathcal K... ...m}} \displaystyle\frac{\ker\bar{b}}{\pi(\ker b)}. \end{array}$$

Theorem 3 (Hutchings [Hu])   All maps in the above chain are isomorphisms. In particular, $(\ker\delta^2)/(\ker\delta)\simeq(\ker\bar{b})/(\pi(\ker b))$.

Proof. Immediate from diagrams (3) and (4). $\Box$

It doesn't look like we've achieved much, but in fact we did, as it seems that $(\ker\bar{b})/(\pi(\ker b))$ is easier to digest than the original space of interest, $(\ker\delta^2)/(\ker\delta)$. The point is that $\ker\bar{b}$ lives fully in the combinatorial realm, being essentially the space of all relations between 4T relations at the symbol level. Similarly, $\pi(\ker b)$ is the space of projections to the symbol level of relation between 4T relations, and hence we have shown

Corollary 1.11   Conjecture 1 is equivalent to the statement ``every relation between 4T relations at the symbol level has a lift to the topological level''.

An obvious approach to proving Conjecture 1 thus emerges:

• Combinatorial step: Find all relations between 4T relations at the symbols level; that is, find a generating set for $\ker\bar{b}$.

• For every relation found in the combinatorial step, show that it lifts to the topological level.

So far, the problem with this approach appears to be in the combinatorial step. There is a conjectural generating set $\bar{\mathcal K}^2_{n-1}$for $\ker\bar{b}$. Every element in $\bar{\mathcal K}^2_{n-1}$ indeed has a lifting to $\ker b$, but we still don't know if $\bar{\mathcal K}^2_{n-1}$indeed generates $\ker\bar{b}$. We state these facts very briefly; more information can be found in [Hu] and in [BS].

Definition 1.12   Define $\bar{\mathcal K}^2_{n-1}$ by

$$\bar{\mathcal K}^2_{n-1} = \operatorname{span}\left\{ \if n... ...{ \input draws/G.tex } \hspace{-1.9mm} \end{array} \right\}.$$

As usual, each graphic in the above formula represents a large number of elements of $\bar{\mathcal K}^2_{n-1}$, obtained from the graphic by the addition of n-3 chords (first graphic), or n-5 chords (second graphic), or n-4 chords (third graphic), or n-2 chords (fourth graphic). Define also $\bar b:\bar{\mathcal K}^2_{n-1}\to\bar{\mathcal K}^1_{n-1}$ by

$$\begin{eqnarray*}\bar b\left( \if ny \smash{\makebox[0pt]{\hspace{-0.5in} \ra... ...pt}{ \input draws/bGExample.tex } \hspace{-1.9mm} \end{array}. \end{eqnarray*}$$

Conjecture 2   The sequence $\bar{\mathcal K}^2_{n-1} \stackrel{\bar b}{\longrightarrow} \bar{\mathcal K}^1_{n-1} \stackrel{\bar b}{\longrightarrow} \bar{\mathcal K}_{n-1}$is exact.

A parallel of Conjecture 2 for braids was proven by Hutchings in [Hu].

Exercise 1.13   Find a space ${\mathcal K}^2_n$ and maps $\delta:{\mathcal K}^2_n\to{\mathcal K}^2_{n-1}$ and $b:{\mathcal K}^2_n\to{\mathcal K}^1_n$ that fit into a commutative diagram,

$$\begin{array}{ccccccc} {\mathcal K}^2_n & \stackrel{\delta}... ... \longrightarrow & 0 \end{array} \qquad\text{(exact rows)},$$

and hence show that the relations in $\ker\bar{b}$ all lift to $\ker b$.

Question 1.14   Is the sequence $\bar{\mathcal K}^2_{n-1} \stackrel{\bar b}{\longrightarrow} \bar{\mathcal K}^1_{n-1} \stackrel{\bar b}{\longrightarrow} \bar{\mathcal K}_{n-1}$ related to Kontsevich's graph cohomology [Ko2]?